Give examples of two functions `f: N->Z` and `g: Z->Z` such that `gof` is injective but `g` is not injective.

Explanation

In f: `N rarrZ, Let f(X) =x`
g: `Z rarr Z,` Let g(x) =|x|
Now g(-2)=g(2) but `-2 ne 2`
`therefore` g is not one - one
Let gof : `N rarr N`
and (gof)(x)= g[f(x)] =g (x)=|x|
Now (gof)(x)=(gof)(y) (where x, y `in` N )
`rArr |x| =|y| `
`rArr x=y " " ( because x, y in N)`
`rArr ` go f is one- one .